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We have the integral domain, for $k \ge 2, k \in \mathbb{Z},$ $$\mathbb{Z}[\sqrt{k} i] = \{m + n \sqrt{k} i : m,n \in \mathbb{Z}\}$$ and a Euclidean function defined by $$v:\mathbb{Z}[\sqrt{k} i] \to \mathbb{N}_0;\;\; v(m + n \sqrt{k} i) = m^2 + kn^2$$ for $m, n \in \mathbb{Z}$. Now let $a_1, \dots ,a_n \in \mathbb{Z}[\sqrt{k} i]$ for some $n \in \mathbb{N}$ and suppose $v(a_1), \dots ,v(a_n)$ are coprime in $\mathbb{Z}$. Show that $a_1, \dots ,a_n$ are coprime in $\mathbb{Z}[\sqrt{k} i]$.

I'm not sure how to proceed with this question. I can write $v(a_1), \dots ,v(a_n)$ as a linear combination of 1, but I don't know how to show the $a_1, \dots ,a_n$ as a similar linear combination.

Please help!

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    $\begingroup$ So you are assuming that the domain is Euclidean with respect to the function $v$? This is not the case in general, if I recall correctly. In any case, if the ring is assumed to be Euclidean, hence a UFD, show that the map $v$ is multiplicative ($v(ab)=v(a)v(b)$ for $a, b \in \mathbb{Z}[\sqrt{k}i]$), and then convince yourself that from that fact, the statement follows. $\endgroup$ – Pavel Čoupek Nov 29 '16 at 17:51
  • $\begingroup$ Do you mean pairwise coprime? $\endgroup$ – Bill Dubuque Nov 29 '16 at 18:08
  • $\begingroup$ @BillDubuque Each $v(a_j)$ is coprime to another $\endgroup$ – Roz Nov 29 '16 at 19:40
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Hint $\,\ v(\alpha) = \alpha\bar \alpha,\ $ so $\,\ 1 = \gcd(v(\alpha), v(\beta)) = \gcd(\alpha\bar \alpha, \beta\bar \beta)\ \Rightarrow\ 1 = \gcd(\alpha,\beta) $

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  • $\begingroup$ $\ldots$ since $\,d\mid \alpha,\beta\,\Rightarrow\, d\mid \alpha\bar\alpha,\beta\bar\beta = v(\alpha),v(\beta)\ $ $\endgroup$ – Bill Dubuque Nov 29 '16 at 18:06

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